To find the intervals of monotonicity for the given functions, we need to follow these steps:
1. **Differentiate the function** to find \( f'(x) \).
2. **Set the derivative equal to zero** to find critical points.
3. **Determine the sign of the derivative** in the intervals defined by the critical points to identify where the function is increasing or decreasing.
Let's solve each part step by step.
### (i) \( f(x) = -x^3 + 6x^2 - 4x - 2 \)
**Step 1: Differentiate the function.**
\[
f'(x) = -3x^2 + 12x - 4
\]
**Step 2: Set the derivative equal to zero.**
\[
-3x^2 + 12x - 4 = 0
\]
Dividing by -1:
\[
3x^2 - 12x + 4 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}
\]
\[
= \frac{12 \pm \sqrt{144 - 48}}{6} = \frac{12 \pm \sqrt{96}}{6} = \frac{12 \pm 4\sqrt{6}}{6} = 2 \pm \frac{2\sqrt{6}}{3}
\]
**Step 3: Determine the sign of \( f'(x) \).**
The critical points are \( x = 2 - \frac{2\sqrt{6}}{3} \) and \( x = 2 + \frac{2\sqrt{6}}{3} \). We will test intervals around these points to determine where the function is increasing or decreasing.
- For \( x < 2 - \frac{2\sqrt{6}}{3} \), choose \( x = 0 \):
\[
f'(0) = -4 < 0 \quad \text{(decreasing)}
\]
- For \( 2 - \frac{2\sqrt{6}}{3} < x < 2 + \frac{2\sqrt{6}}{3} \), choose \( x = 2 \):
\[
f'(2) = -3(2)^2 + 12(2) - 4 = 0 \quad \text{(check around)}
\]
- For \( x > 2 + \frac{2\sqrt{6}}{3} \), choose \( x = 3 \):
\[
f'(3) = -3(3)^2 + 12(3) - 4 = 2 > 0 \quad \text{(increasing)}
\]
Thus, the function is:
- Decreasing on \( (-\infty, 2 - \frac{2\sqrt{6}}{3}) \)
- Increasing on \( (2 + \frac{2\sqrt{6}}{3}, \infty) \)
### (ii) \( f(x) = x + \frac{1}{x+1} \)
**Step 1: Differentiate the function.**
\[
f'(x) = 1 - \frac{1}{(x+1)^2}
\]
**Step 2: Set the derivative equal to zero.**
\[
1 - \frac{1}{(x+1)^2} = 0 \implies (x+1)^2 = 1 \implies x + 1 = \pm 1
\]
Thus, \( x = 0 \) or \( x = -2 \).
**Step 3: Determine the sign of \( f'(x) \).**
- For \( x < -2 \), choose \( x = -3 \):
\[
f'(-3) = 1 - \frac{1}{(-2)^2} = 1 - \frac{1}{4} = \frac{3}{4} > 0 \quad \text{(increasing)}
\]
- For \( -2 < x < 0 \), choose \( x = -1 \):
\[
f'(-1) = 1 - 1 = 0 \quad \text{(check)}
\]
- For \( x > 0 \), choose \( x = 1 \):
\[
f'(1) = 1 - \frac{1}{(2)^2} = 1 - \frac{1}{4} = \frac{3}{4} > 0 \quad \text{(increasing)}
\]
Thus, the function is:
- Increasing on \( (-\infty, -2) \) and \( (0, \infty) \)
- Decreasing on \( (-2, 0) \)
### (iii) \( f(x) = x e^{x - x^2} \)
**Step 1: Differentiate the function using the product rule.**
\[
f'(x) = e^{x-x^2} + x e^{x-x^2} (1 - 2x)
\]
Factoring out \( e^{x-x^2} \):
\[
f'(x) = e^{x-x^2} (1 + x(1 - 2x))
\]
**Step 2: Set the derivative equal to zero.**
\[
1 + x(1 - 2x) = 0 \implies 1 + x - 2x^2 = 0 \implies 2x^2 - x - 1 = 0
\]
Using the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
\]
Thus, \( x = 1 \) or \( x = -\frac{1}{2} \).
**Step 3: Determine the sign of \( f'(x) \).**
- For \( x < -\frac{1}{2} \), choose \( x = -1 \):
\[
f'(-1) = e^{-1} (1 - 2(-1)^2) = e^{-1} (1 - 2) < 0 \quad \text{(decreasing)}
\]
- For \( -\frac{1}{2} < x < 1 \), choose \( x = 0 \):
\[
f'(0) = e^0 (1 + 0) > 0 \quad \text{(increasing)}
\]
- For \( x > 1 \), choose \( x = 2 \):
\[
f'(2) = e^{2-4} (1 + 2(1 - 4)) < 0 \quad \text{(decreasing)}
\]
Thus, the function is:
- Decreasing on \( (-\infty, -\frac{1}{2}) \) and \( (1, \infty) \)
- Increasing on \( (-\frac{1}{2}, 1) \)
### (iv) \( f(x) = x - \cos x \)
**Step 1: Differentiate the function.**
\[
f'(x) = 1 + \sin x
\]
**Step 2: Set the derivative equal to zero.**
\[
1 + \sin x = 0 \implies \sin x = -1
\]
This occurs at \( x = \frac{3\pi}{2} + 2k\pi \) for \( k \in \mathbb{Z} \).
**Step 3: Determine the sign of \( f'(x) \).**
Since \( \sin x \) oscillates between -1 and 1, \( f'(x) \) will always be greater than 0 except at the points where \( \sin x = -1 \). Thus, \( f'(x) \) is positive everywhere except at those specific points.
Thus, the function is:
- Increasing on \( (-\infty, \frac{3\pi}{2}) \) and \( (\frac{3\pi}{2}, \infty) \)
### Summary of Intervals of Monotonicity:
1. \( f(x) = -x^3 + 6x^2 - 4x - 2 \)
- Decreasing: \( (-\infty, 2 - \frac{2\sqrt{6}}{3}) \)
- Increasing: \( (2 + \frac{2\sqrt{6}}{3}, \infty) \)
2. \( f(x) = x + \frac{1}{x+1} \)
- Increasing: \( (-\infty, -2) \) and \( (0, \infty) \)
- Decreasing: \( (-2, 0) \)
3. \( f(x) = x e^{x - x^2} \)
- Decreasing: \( (-\infty, -\frac{1}{2}) \) and \( (1, \infty) \)
- Increasing: \( (-\frac{1}{2}, 1) \)
4. \( f(x) = x - \cos x \)
- Increasing: \( (-\infty, \infty) \) except at \( x = \frac{3\pi}{2} + 2k\pi \)
